f(x)=(x/2)^2

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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simplify solve for inverse tangent line

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f(x)=(x2 )²

Solution

Domain: −∞ <x<∞

Range: f(x)≥ 0

X Intercepts: (0, 0), Y Intercepts: (0, 0)

Vertex: Minimum (0, 0)

Inverse: 2√x, −2√x

Interval Notation

Domain: (−∞ , ∞ )

Range: [0, ∞ )

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Solution steps

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Domain of (x2 )² : −∞ <x<∞

Range of (x2 )²: f(x)≥ 0

Axis interception points of (x2 )²: X Intercepts: (0, 0), Y Intercepts: (0, 0)

Vertex of (x2 )²: Minimum (0, 0)

Inverse of (x2 )²: 2√x, −2√x

Solve instead

ddx ((x2 )²)

parity (x2 )²

symmetry (x2 )²

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

f(x)=(x/2)^2

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